11. An "ice calorimeter" can be used to determine the specific heat capacity of a metal. Suppose you heat a 50.0 gram piece of an unknown metal to 99.8 °C and then drop it onto ice. When the metal's temperature has dropped to 0.0 °C, it is found that 3.54 grams of ice had melted. What is the specific heat capacity of the metal? Assume no heat lost to the surroundings: Heat released by metal = Heat absorbed by ICE %3D Heat absorbed by ICE = mAHfus ; Heat released by metal = -mcAT qice + qmetal = 0 = (3.54 g)(333 J/g)+ (50.0 g)(Cmetal)(0.0°C – 99.8°C) %3D %3D 0 = 1178.82 J- (4990 g °C)cmetal > Cmetal = 0.236 J/(g °C) %3D
11. An "ice calorimeter" can be used to determine the specific heat capacity of a metal. Suppose you heat a 50.0 gram piece of an unknown metal to 99.8 °C and then drop it onto ice. When the metal's temperature has dropped to 0.0 °C, it is found that 3.54 grams of ice had melted. What is the specific heat capacity of the metal? Assume no heat lost to the surroundings: Heat released by metal = Heat absorbed by ICE %3D Heat absorbed by ICE = mAHfus ; Heat released by metal = -mcAT qice + qmetal = 0 = (3.54 g)(333 J/g)+ (50.0 g)(Cmetal)(0.0°C – 99.8°C) %3D %3D 0 = 1178.82 J- (4990 g °C)cmetal > Cmetal = 0.236 J/(g °C) %3D
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter6: Thermochemistry
Section: Chapter Questions
Problem 64E: A 110.-g sample of copper (specific heat capacity = 0.20 J/C g) is heated to 82.4C and then placed...
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For the part circled in pencil, where did this number come from? For the part circled in yellow, what is this equation referred to as or representative of?
Transcribed Image Text:11. An "ice calorimeter" can be used to determine the specific heat capacity of a metal.
Suppose you heat a 50.0 gram piece of an unknown metal to 99.8 °C and then drop it onto
ice. When the metal's temperature has dropped to 0.0 °C, it is found that 3.54 grams of ice
had melted. What is the specific heat capacity of the metal?
Assume no heat lost to the surroundings: Heat released by metal = Heat absorbed by ICE
%3D
Heat absorbed by ICE = mAHfus ; Heat released by metal = -mcAT
6.
qice + qmetal = 0 = (3.54 g)(333 J/g) + (50.0 g)(Cmetal)(0.0°C – 99.8°C)
%3D
%3D
0 = 1178.82 J – (4990 g °C)Cmetal → Cmetal = 0.236 J/(g °C)
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